stMale)Geiser C. Challco geiser@alumni.usp.br
env <- "stMale"
gender <- "men"
to_remove <- c('S11')
sub.groups <- c("age","ed.level","intervention","age:intervention",
"ed.level:intervention","age:ed.level:intervention")dat <- read_excel("../data/data-without-outliers.xlsx", sheet = "fss-env.gender-descriptive")
dat <- dat[!dat$study %in% to_remove, ]
leg <- read_excel("../data/data-without-outliers.xlsx", sheet = "legend")## New names:
## • `` -> `...10`leg <- leg[!leg$study %in% to_remove, ]
idx.e <- which(dat$env == env & dat$gender == gender)
idx.c <- which(dat$env == "control" & dat$gender == gender)
data <- data.frame(
study = dat$study[idx.c],
n.e = dat$N[idx.e], mean.e = dat$M.emms[idx.e], sd.e = dat$SD.emms[idx.e],
n.c = dat$N[idx.c], mean.c = dat$M.emms[idx.c], sd.c = dat$SD.emms[idx.c]
)
for (cgroups in strsplit(sub.groups,":")) {
data[[paste0(cgroups, collapse = ":")]] <- sapply(data$study, FUN = function(x) {
paste0(sapply(cgroups, FUN = function(namecol) leg[[namecol]][which(x == leg$study)]), collapse = ":")
})
}
data[["lbl"]] <- sapply(data$study, FUN = function(x) leg$Note[which(x == leg$study)])m.cont <- metacont(
n.e = n.e, mean.e = mean.e, sd.e = sd.e, n.c = n.c, mean.c = mean.c, sd.c = sd.c,
studlab = lbl, data = data, sm = "SMD", method.smd = "Hedges",
fixed = F, random = T, method.tau = "REML", hakn = T, title = paste("Flow state for",gender,"in",env)
)
summary(m.cont)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random)
## S1 0.3993 [-0.4347; 1.2333] 7.3
## S2 0.3329 [-0.4228; 1.0886] 8.9
## S3 0.0654 [-0.7378; 0.8686] 7.9
## S4 -0.0889 [-0.9154; 0.7375] 7.4
## S5 -0.1134 [-0.6940; 0.4673] 15.1
## S6 0.2084 [-0.3671; 0.7838] 15.4
## S7 0.1350 [-0.3943; 0.6643] 18.2
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.cont, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.sg4sub <- update.meta(m.cont, subgroup = age, random = T, fixed = F)
summary(m.sg4sub)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random) age
## S1 0.3993 [-0.4347; 1.2333] 7.3 adolescent
## S2 0.3329 [-0.4228; 1.0886] 8.9 adolescent
## S3 0.0654 [-0.7378; 0.8686] 7.9 adolescent
## S4 -0.0889 [-0.9154; 0.7375] 7.4 adult
## S5 -0.1134 [-0.6940; 0.4673] 15.1 adult
## S6 0.2084 [-0.3671; 0.7838] 15.4 adult
## S7 0.1350 [-0.3943; 0.6643] 18.2 adult
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8 adolescent
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Results for subgroups (random effects model):
## k SMD 95%-CI tau^2 tau Q I^2
## age = adolescent 4 0.1135 [-0.2502; 0.4772] 0 0 1.30 0.0%
## age = adult 4 0.0585 [-0.1940; 0.3110] 0 0 0.80 0.0%
##
## Test for subgroup differences (random effects model):
## Q d.f. p-value
## Between groups 0.16 1 0.6926
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.sg4sub, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.sg4sub <- update.meta(m.cont, subgroup = ed.level, random = T, fixed = F)
summary(m.sg4sub)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random) ed.level
## S1 0.3993 [-0.4347; 1.2333] 7.3 upper-secundary
## S2 0.3329 [-0.4228; 1.0886] 8.9 upper-secundary
## S3 0.0654 [-0.7378; 0.8686] 7.9 upper-secundary
## S4 -0.0889 [-0.9154; 0.7375] 7.4 higher-education
## S5 -0.1134 [-0.6940; 0.4673] 15.1 higher-education
## S6 0.2084 [-0.3671; 0.7838] 15.4 higher-education
## S7 0.1350 [-0.3943; 0.6643] 18.2 unknown
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8 upper-secundary
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Results for subgroups (random effects model):
## k SMD 95%-CI tau^2 tau Q I^2
## ed.level = upper-secundary 4 0.1135 [-0.2502; 0.4772] 0 0 1.30 0.0%
## ed.level = higher-education 3 0.0219 [-0.4475; 0.4912] 0 0 0.68 0.0%
## ed.level = unknown 1 0.1350 [-0.3943; 0.6643] -- -- 0.00 --
##
## Test for subgroup differences (random effects model):
## Q d.f. p-value
## Between groups 0.40 2 0.8198
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.sg4sub, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.sg4sub <- update.meta(m.cont, subgroup = intervention, random = T, fixed = F)
summary(m.sg4sub)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random)
## S1 0.3993 [-0.4347; 1.2333] 7.3
## S2 0.3329 [-0.4228; 1.0886] 8.9
## S3 0.0654 [-0.7378; 0.8686] 7.9
## S4 -0.0889 [-0.9154; 0.7375] 7.4
## S5 -0.1134 [-0.6940; 0.4673] 15.1
## S6 0.2084 [-0.3671; 0.7838] 15.4
## S7 0.1350 [-0.3943; 0.6643] 18.2
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8
## intervention
## S1 Gender-stereotype color, ranking, badges, and avatar
## S2 Gender-stereotype color, ranking, badges, and avatar
## S3 Gender-stereotype color, ranking, badges, and avatar
## S4 Gender-stereotype color, ranking, badges, and avatar
## S5 Gender-stereotype color, ranking, badges, and avatar
## S6 Gender-stereotype color, ranking, badges, and avatar
## S7 Gender-stereotype color, ranking, badges, and avatar
## S10: Only use prompt msgs Gender-stereotyped motivational message prompts
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Results for subgroups (random effects model):
## k SMD 95%-CI tau^2 tau Q I^2
## intervention = Gender-stereotype color, rankin ... 7 0.1208 [-0.0473; 0.2888] 0 0 1.71 0.0%
## intervention = Gender-stereotyped motivational ... 1 -0.0713 [-0.5778; 0.4352] -- -- 0.00 --
##
## Test for subgroup differences (random effects model):
## Q d.f. p-value
## Between groups 0.52 1 0.4725
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.sg4sub, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.sg4sub <- update.meta(m.cont, subgroup = `age:intervention`, random = T, fixed = F)
summary(m.sg4sub)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random)
## S1 0.3993 [-0.4347; 1.2333] 7.3
## S2 0.3329 [-0.4228; 1.0886] 8.9
## S3 0.0654 [-0.7378; 0.8686] 7.9
## S4 -0.0889 [-0.9154; 0.7375] 7.4
## S5 -0.1134 [-0.6940; 0.4673] 15.1
## S6 0.2084 [-0.3671; 0.7838] 15.4
## S7 0.1350 [-0.3943; 0.6643] 18.2
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8
## age:intervention
## S1 adolescent:Gender-stereotype color, ranking, badges, and avatar
## S2 adolescent:Gender-stereotype color, ranking, badges, and avatar
## S3 adolescent:Gender-stereotype color, ranking, badges, and avatar
## S4 adult:Gender-stereotype color, ranking, badges, and avatar
## S5 adult:Gender-stereotype color, ranking, badges, and avatar
## S6 adult:Gender-stereotype color, ranking, badges, and avatar
## S7 adult:Gender-stereotype color, ranking, badges, and avatar
## S10: Only use prompt msgs adolescent:Gender-stereotyped motivational message prompts
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Results for subgroups (random effects model):
## k SMD 95%-CI tau^2 tau Q I^2
## age:intervention = adolescent:Gender-stereotype co ... 3 0.2655 [-0.1669; 0.6980] 0 0 0.37 0.0%
## age:intervention = adult:Gender-stereotype color, ... 4 0.0585 [-0.1940; 0.3110] 0 0 0.80 0.0%
## age:intervention = adolescent:Gender-stereotyped m ... 1 -0.0713 [-0.5778; 0.4352] -- -- 0.00 --
##
## Test for subgroup differences (random effects model):
## Q d.f. p-value
## Between groups 3.23 2 0.1985
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.sg4sub, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.sg4sub <- update.meta(m.cont, subgroup = `ed.level:intervention`, random = T, fixed = F)
summary(m.sg4sub)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random)
## S1 0.3993 [-0.4347; 1.2333] 7.3
## S2 0.3329 [-0.4228; 1.0886] 8.9
## S3 0.0654 [-0.7378; 0.8686] 7.9
## S4 -0.0889 [-0.9154; 0.7375] 7.4
## S5 -0.1134 [-0.6940; 0.4673] 15.1
## S6 0.2084 [-0.3671; 0.7838] 15.4
## S7 0.1350 [-0.3943; 0.6643] 18.2
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8
## ed.level:intervention
## S1 upper-secundary:Gender-stereotype color, ranking, badges, and avatar
## S2 upper-secundary:Gender-stereotype color, ranking, badges, and avatar
## S3 upper-secundary:Gender-stereotype color, ranking, badges, and avatar
## S4 higher-education:Gender-stereotype color, ranking, badges, and avatar
## S5 higher-education:Gender-stereotype color, ranking, badges, and avatar
## S6 higher-education:Gender-stereotype color, ranking, badges, and avatar
## S7 unknown:Gender-stereotype color, ranking, badges, and avatar
## S10: Only use prompt msgs upper-secundary:Gender-stereotyped motivational message prompts
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Results for subgroups (random effects model):
## k SMD 95%-CI tau^2 tau Q
## ed.level:intervention = upper-secundary:Gender-stereoty ... 3 0.2655 [-0.1669; 0.6980] 0 0 0.37
## ed.level:intervention = higher-education:Gender-stereot ... 3 0.0219 [-0.4475; 0.4912] 0 0 0.68
## ed.level:intervention = unknown:Gender-stereotype color ... 1 0.1350 [-0.3943; 0.6643] -- -- 0.00
## ed.level:intervention = upper-secundary:Gender-stereoty ... 1 -0.0713 [-0.5778; 0.4352] -- -- 0.00
## I^2
## ed.level:intervention = upper-secundary:Gender-stereoty ... 0.0%
## ed.level:intervention = higher-education:Gender-stereot ... 0.0%
## ed.level:intervention = unknown:Gender-stereotype color ... --
## ed.level:intervention = upper-secundary:Gender-stereoty ... --
##
## Test for subgroup differences (random effects model):
## Q d.f. p-value
## Between groups 3.40 3 0.3340
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.sg4sub, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.sg4sub <- update.meta(m.cont, subgroup = `age:ed.level:intervention`, random = T, fixed = F)
summary(m.sg4sub)## Review: Flow state for men in stMale
##
## SMD 95%-CI %W(random)
## S1 0.3993 [-0.4347; 1.2333] 7.3
## S2 0.3329 [-0.4228; 1.0886] 8.9
## S3 0.0654 [-0.7378; 0.8686] 7.9
## S4 -0.0889 [-0.9154; 0.7375] 7.4
## S5 -0.1134 [-0.6940; 0.4673] 15.1
## S6 0.2084 [-0.3671; 0.7838] 15.4
## S7 0.1350 [-0.3943; 0.6643] 18.2
## S10: Only use prompt msgs -0.0713 [-0.5778; 0.4352] 19.8
## age:ed.level:intervention
## S1 adolescent:upper-secundary:Gender-stereotype color, ranking, badges, and avatar
## S2 adolescent:upper-secundary:Gender-stereotype color, ranking, badges, and avatar
## S3 adolescent:upper-secundary:Gender-stereotype color, ranking, badges, and avatar
## S4 adult:higher-education:Gender-stereotype color, ranking, badges, and avatar
## S5 adult:higher-education:Gender-stereotype color, ranking, badges, and avatar
## S6 adult:higher-education:Gender-stereotype color, ranking, badges, and avatar
## S7 adult:unknown:Gender-stereotype color, ranking, badges, and avatar
## S10: Only use prompt msgs adolescent:upper-secundary:Gender-stereotyped motivational message prompts
##
## Number of studies combined: k = 8
## Number of observations: o = 318
##
## SMD 95%-CI t p-value
## Random effects model 0.0827 [-0.0683; 0.2337] 1.29 0.2366
##
## Quantifying heterogeneity:
## tau^2 = 0 [0.0000; 0.0309]; tau = 0 [0.0000; 0.1759]
## I^2 = 0.0% [0.0%; 67.6%]; H = 1.00 [1.00; 1.76]
##
## Test of heterogeneity:
## Q d.f. p-value
## 2.16 7 0.9507
##
## Results for subgroups (random effects model):
## k SMD 95%-CI tau^2 tau
## age:ed.level:intervention = adolescent:upper-secundary:Gend ... 3 0.2655 [-0.1669; 0.6980] 0 0
## age:ed.level:intervention = adult:higher-education:Gender-s ... 3 0.0219 [-0.4475; 0.4912] 0 0
## age:ed.level:intervention = adult:unknown:Gender-stereotype ... 1 0.1350 [-0.3943; 0.6643] -- --
## age:ed.level:intervention = adolescent:upper-secundary:Gend ... 1 -0.0713 [-0.5778; 0.4352] -- --
## Q I^2
## age:ed.level:intervention = adolescent:upper-secundary:Gend ... 0.37 0.0%
## age:ed.level:intervention = adult:higher-education:Gender-s ... 0.68 0.0%
## age:ed.level:intervention = adult:unknown:Gender-stereotype ... 0.00 --
## age:ed.level:intervention = adolescent:upper-secundary:Gend ... 0.00 --
##
## Test for subgroup differences (random effects model):
## Q d.f. p-value
## Between groups 3.40 3 0.3340
##
## Details on meta-analytical method:
## - Inverse variance method
## - Restricted maximum-likelihood estimator for tau^2
## - Q-profile method for confidence interval of tau^2 and tau
## - Hartung-Knapp adjustment for random effects model
## - Hedges' g (bias corrected standardised mean difference; using exact formulae)forest(m.sg4sub, digits=2, digits.sd = 2, test.overall = T, label.e = paste0(gender,':',env))
m.cont <- update.meta(m.cont, studlab = data$study)
summary(eggers.test(x = m.cont))## Eggers' test of the intercept
## =============================
##
## intercept 95% CI t p
## 1.02 -0.96 - 3 1.012 0.35
##
## Eggers' test does not indicate the presence of funnel plot asymmetry.funnel(m.cont, xlab = "Hedges' g", studlab = T, legend=T, addtau2 = T)